Question: Let $h(x)=-2x^5+10x^4$. For what values of $x$ does the graph of $h$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=3$ (Choice C) C $x=5$ (Choice D) D $h$ has no points of inflection.
Answer: We can find the inflection points of the graph of $h$ by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=-40x^2(x-3)$. $h''(x)=0$ for $x=0,3$. Since $h''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=0$ and $x=3$. Our possible inflection points divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $(-\infty,0)$ $(0,3)$ $(3,\infty)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $(-\infty,0)$ $x=-1$ $h''(-1)=160>0$ $h$ is concave up $\cup$ $(0,3)$ $x=2$ $h''(2)=160>0$ $h$ is concave up $\cup$ $(3,\infty)$ $x=4$ $h''(4)=-640<0$ $h$ is concave down $\cap$ We can see that the graph of $h$ changes concavity at $x=3$. In conclusion, $h$ has a point of inflection at $x=3$.